Finding the remainder when $5^{55}+3^{55}$ is divided by $16$

$$5^{55}+3^{55}\equiv(125\times625^{13})+(27\times81^{13})\pmod {16}$$ $$\equiv 125\times 1+27\times 1\pmod{16}$$ $$\equiv 8\pmod {16}$$

Notice that $5 = 4+1$ and $3=4-1$. Then by the binomial theorem we have

$$5^{55}+3^{55} = (4+1)^{55}+(4-1)^{55} =$$

$$\sum_{k=0}^{55} {55 \choose k} 4^{55-k}1^{k} + \sum_{k=0}^{55} {55 \choose k} 4^{55-k}(-1)^{k}.$$

Now every term in sight has $4^2 = 16$ except the last two in each sum. So the above equals

$$55(4) +1 + 55(4) -1 = 8(55) = 8(54) +8.$$

Since $8(54)$ is a multiple of $16$, the answer is $8$.

Let $a_n = 5^{n}+3^{n}$. Then $a_{n+2} = 8a_{n+1} -15 a_n$ because $5$ and $3$ are roots of $x^2=8x-15$. Mod $16$ we get $$ 2,8,2,8,\dots $$ that is, the sequence is periodic of period $2$. This can be proved by induction. Therefore, $a_{55} \equiv a_1 = 8$.